Friday

August 29, 2014

August 29, 2014

Posted by **sied** on Thursday, October 10, 2013 at 7:44pm.

What is the work done by the friction force on the block while it is sliding down the inclined plane? Express your answer in terms of , , , , , and (enter theta for , mu_1 for , and mu_2 for ).

unanswered

Answer:

(b) What is the work done by the gravitational force on the block while it is sliding down the inclined plane? Express your answer in terms of , , , , , and (enter theta for , mu_1 for , and mu_2 for ).

unanswered

Answer:

(c) What is the kinetic energy of the block just at the bottom of the inclined plane? Express your answer in terms of , , , , , and (enter theta for , mu_1 for , and mu_2 for ).

d) After leaving the incline, the block slides along the rough surface until it comes to rest. How far has it traveled? Express your answer in terms of , , , , , and (enter theta for , mu_1 for , and mu_2 for ).

- Physics -
**Count Iblis**, Thursday, October 10, 2013 at 10:14pmThe height is L sin(theta). The decrease in the gravitational potential energy is thus m g L sin(theta).

The magnitude of the component of the gravitational force orthogonal to the incline is m g cos(theta), therefore the normal force is equal to

m g cos(theta). The friction force is thus equal to mu_1 m g cos(theta), the work done by the friction force is thus:

- mu_1 m g cos(theta) L

the minus sign coming from the fact that the force has a direction oposite to the displacement.

The kinetic energy at the bottom of the incline is thus given by:

m g L [sin(theta) - mu_1 cos(theta)]

Then there is a big mistake in the problem, because it assumes that the block will start to slide on the rough horizontal surface with with an initial velocity that corresponds to the above kinetic energy, but this wrong. I'll explain that later.

If the kinetic energy at the start of the horizntal rough surface is E, then you obviously have that the distance d it will travel satisfies:

mu_2 m g d = E

So, d = E/(mu_2 m g).

Now, you can't take E equal to

m g L [sin(theta) - mu_1 cos(theta)]

because the block has to change direction which requires an extra normal force and therefore additional friction forces. To calculate this effect, let's assume that the change in direction happens on the surface with coefficient of friction mu_1 over a very short distance at the ground level.

Then what happens is that the component of the block in the vertical direction will have to vanish when the incline levels off and the block is about to enter the rough surface. The integral of the component of the normal force in the vertical direction over time during the change of direction is thus equal to p sin(theta) where p is the magnitude of the initial momentum.

The friction force is always orthogonal to the normal force, and equal to mu_1 times that normal force. Therefore the component of the friction force in the horizontal direction is - mu_1 times the component if the normal force in the vertical direction, the integral over time during the change in direction is thus equal to

-mu_1 p sin(theta) and this is then the change in the component of the momentum in the horizontal direction.

The initial kinetic energy was:

E1 =

m g L [sin(theta) - mu_1 cos(theta)]

This is also given in terms of the momentum as:

E1 = p^2/(2m)

therefore:

|p| = sqrt(2 m E1)

the initial horizontal component is thus:

|p| cos(theta) = sqrt(2 m E1) cos(theta)

The final horizontal component is thus:

sqrt(2 m E1)*

[cos(theta) - mu_1 sin(theta)]

and this is the total momentum because the vertical component will have vanished as the incline levels off.

The kinetic energy at the start of the rough surface is thus:

E = E1 [cos(theta) - mu_1 sin(theta)]

The distance the block will slide is thus given by:

d = E/(mu_2 m g) =

L/mu_2 [cos(theta) - mu_1 sin(theta)]*

[sin(theta) - mu_1 cos(theta)] =

[1/2(1+mu_1^2)sin(2 theta)-mu_1]L/mu_2

- Physics -
**Count Iblis**, Friday, October 11, 2013 at 10:19amCorrection of the last part:

The kinetic energy at the start of the rough surface is thus:

E = E1 [cos(theta) - mu_1 sin(theta)]^2

The distance the block will slide is thus given by:

d = E/(mu_2 m g) =

L/mu_2 [cos(theta) - mu_1 sin(theta)]^2* [sin(theta) - mu_1 cos(theta)] =

[1/2(1+mu_1^2)sin(2 theta)-mu_1]*

[cos(theta) - mu_1 sin(theta)] L/mu_2

- Physics -
**dude**, Tuesday, October 15, 2013 at 3:33pmu r a genious man..!

**Related Questions**

General Physics - The 5 kg block in the figure slides down a frictionless curved...

Physics - A 4.3-kg block slides down an inclined plane that makes an angle of 26...

Physics - A 4.6-kg block slides down an inclined plane that makes an angle of 26...

physics - A 5.4 kg block slides down an inclined plane that makes an angle of 38...

Physics - A 5.1 kg block slides down an inclined plane that makes an angle of 26...

physics - An block of mass m , starting from rest, slides down an inclined plane...

physics - An block of mass m , starting from rest, slides down an inclined plane...

physics - A block of mass 12.0 kg slides from rest down a frictionless 35.0° ...

physics - The 2 kg block in Figure 7-25 slides down a frictionless curved ramp, ...

ap physics - Starting from rest, a 10.2 kg block slides 2.90 m down a ...